# Details in existence of free groups proof (Clara Loeh,pg-22,23)

I have a few doubts in the proof of existence of free groups given by Clara Loeh in page-22,23 of her Geometric Group theory book.

Theorem 2.2.7: Let S be a set. Then there exists a group freely generated by S. (By the previous proposition, this group is unique up to isomorphism.)

1. When the equivalence is defined in page-22, is it implicitly assumed that $$s$$ and $$\hat{s}$$ are inverses?

$$\forall_{x,y \in A^*} \forall_{s \in S} xs \hat{s}y \sim y xy$$ $$\forall_{x,y \in A^*} \forall_{s \in S} x \hat{s} s y \sim xy$$

1. In page-23, Clara defines a function which help us define inverse for general words. Here is what she gave exactly:

Inductively (over the length of sequences), we define a map $$I:A^* \to A^*$$ by $$I(\epsilon) := \epsilon$$ and

1. $$I(sx) :=I(x) \hat{s}$$

2. $$I( \hat{s} x) : = I(x) s$$

In the above why did she write 1. and 2.? Shouldn't just one of the equations be enough? My thinking is that in $$A^*$$ , $$s$$ and $$\hat{s}$$ are just elements in the set $$A^*$$.

Secondly, I am not sure if I understand this function in itself. Suppose I have a word $$abc$$ made up of letters $$a,b,c$$ then would it be correct to write $$I(abc)= I(bc) \hat{a}$$..? and then push it down till $$I(abc)=\hat{c} \hat{b} \hat{a}$$?

1. In page-23 (again), I am confused at the part where she defines $$\varphi^*$$ which map from the set to words to a group. The idea can be understood by the diagram:

For every group G and every map$$\varphi: S −→ G$$ there is a unique group homomorphism $$\varphi: F(S) −→ G$$ such that $$\overline{\varphi} ◦ i = \varphi$$. Given $$\varphi$$, we construct a map

$$\phi^*: A^* \to G$$

Inductively by:

1. $$\epsilon \to e$$

2. $$sx \to \varphi(s) \cdot \varphi^*(x)$$

3. $$\hat{s} x \to ( \varphi(s))^{-1} \cdot \varphi^*(x)$$

Again I am confused, why we need both two and three..? I think one is enough for similar reason said before.

## Note:

1. Free groups are defined by their universal property in the book.

For your first question: the proof of theorem $$2.2.7$$ explicitly explains what $$\hat{s}$$ is: it's an additional disjoint copy of $$S$$ whose elements play the roles of inverses. In fact, it's the very first paragraph of the proof.
For your second question: since $$s\in S$$ and $$\hat s \in S$$ are different (and belong to different, disjoint copies of $$S$$) we need to define how they both operate on the group element $$x$$, which is why both points $$1$$ and $$2$$ are given. You can indeed go from $$I(abc)$$ to $$\hat{c}\hat b\hat a$$ and this is also the expected right-inverse of the word $$abc$$ in a group.
And for your last question the answer is again that this explicitly connects the nominal inverse $$\hat s$$ with the inverse of the homomorphism $$\varphi$$ so both need to be defined.
• If the group is not commutative then knowing what $x\hat s$ is tells you nothing about $\hat s x$. So you have to define both terms separately. Nominal inverse means "that thing we are calling an inverse" because: $S$ is set and sets don't have inverses. But if we can treat $S$ as a group, then it does, because groups are required to have inverses. Since groups are also closed, all the inverses $\hat s$ of elements $s\in S$ must also be in $S$, so we can take a second copy of $S$ and treat them as in the inverses. We're calling them inverses, but at this stage we haven't shown they are Apr 12 at 11:38
• I don't really understand what you're writing there I'm afraid. $\hat s x$ and $x \hat s$ are different things, as Ms. Löh defines, and are on equal footing inasmuch as they are both words. Apr 12 at 11:55
• Ah ok! I'm sorry, that's completely my fault; points one and two are not about commutativity (I won't bore you with the connection that made me write that); I've updated the answer to reflect that. The issue is that $s$ and $\hat s$ belong to different copies of $S$ so we need to specify how they both behave with respect to $I$. Apr 12 at 14:23
• Could you explain right side of third eqtn? I am confused at $( \varphi(s))^{-1}$ Apr 12 at 14:41
• If you have more questions you should really ask them as a new question and not keep adding them in comments. However: the point of $3.$ is to establish that $\phi ^*$ is a homomorphism and that it interacts with $\hat s$ by inverses (because we intend $\hat s$ to be the inverse of $s$). From $\epsilon = \phi(\varepsilon) = \phi(\hat s s) = \phi(\hat s) \phi(s) \Rightarrow (\phi(s))^{-1} = \phi(\hat s)$ (writing the homomorphism multiplicatively) being what we want we obtain $3$. Apr 13 at 6:46